R S Aggarwal Solutions for Class 10 Maths Chapter 6 exercise 6c is available here. These R S Aggarwal solutions are prepared by subject experts to help students clear their doubts. The concepts are explained using step by step approach. This exercise is based on topic area of a triangle using vertex coordinates. Students can download Class 10 Maths Chapter 6 R S Aggarwal Solutions and start practicing offline.

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Exercise 6A Solutions: 20 Questions (20 Short Answers)

Exercise 6B Solutions: 20 Questions (20 Short Answers)

Exercise 6D Solutions: 17 Questions (12 Short Answers & 5 Long Answers )

**Exercise 6C Page No: 336**

**Question 1: Find the area of âˆ†ABC whose vertices are:**

**(i) A (1, 2), B (-2, 3) and C (-3, -4)**

**(ii) A (-5, 7), B (-4, -5) and C (4, 5)**

**(iii) A (3, 8), B (-4, 2) and C (5, -1)**

**(iv) A (10, -6), B (2, 5) and C (-1, 3)**

**Solution:**

Area of âˆ†ABC whose vertices are is (x_1,y_1 ),(x_2,y_2 )and (x_3,y_3 ) are

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)(i) In âˆ†ABC, vertices are A (1, 2), B (-2, 3) and C (-3, -4)

Area of triangle = 1/2(1(â€“2 + 3)â€“2(â€“4â€“2)â€“3(2â€“3))

= 1/2(1 + 12 + 3)

= 8 sq units

(ii) A (-5, 7), B (-4, -5) and C (4, 5)

Area of triangle = 1/2(â€“5(â€“5â€“5)â€“4(5â€“7) + 4(7 + 5))

= 1/2(â€“50 + 8 + 48)

= 5 sq units

(iii) A (3, 8), B (-4, 2) and C (5, -1)

Area of triangle = 1/2(3(2 + 1)â€“4(â€“1â€“8) + 5(8â€“2))

= 1/2(9 + 36 + 30)

= 1/2(75)

= 37.5 sq units

(iv) A (10, -6), B (2, 5) and C (-1, 3)

Area of triangle = 1/2(10(5â€“3) + 2(3 + 6)â€“1(â€“6â€“5))

= 1/2(20 + 18 + 11)

= 1/2(49)

= 24.5 sq units

**Question 2: Find the area of quadrilateral ABCD whose vertices are A (3, -1), B (9, -5), C (14, 0) and D (9, 19). **

**Solution: **Vertices of quadrilateral ABCD are A(3,-1), B(9, -5), C (14, 0) and D(9, 19)

Construction: Join diagonal AC.

We know that:

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)**Question 3: Find the area of quadrilateral PQRS whose vertices are P (-5, -3), Q (-4, -6), R (2, -3) and S (1, 2). **

**Solution:**

Given: PQRS is a quadrilateral whose vertices are P(-5, -3), Q(-4, -6), R(2, -3) and S(1, 2)

Construction: Join PR

We know that:

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)Now, Area of quadrilateral PQRS = Area of triangle PQR + Area of triangle PSR

= 21/2 + 35/2

= 28 sq. units

**Question 4: Find the area of quadrilateral ABCD whose vertices are A (-3, -1), B (-2, -4), C (4, -1) and D (3, 4). **

**Solution:**

Given: ABCD is a quadrilateral whose vertices are A (-3, -1), B (-2, -4), C (4, -1) and D (3, 4).

By Joining AC, we get two triangles ABC and ADC

**Question 5: If A (-7, 5), B (-6, -7), C (-3, -8) and D (2, 3) are the vertices of a quadrilateral ABCD then find the area of the quadrilateral.**

**Solution:**

**Question 6: Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5). **

**Solution:**

Given: A triangle whose vertices are A(2, 1), B(4, 3) and C(2, 5)

Let D, E and F are the midpoints of the sides CB, CA and AB respectively of âˆ†ABC, as shown in the below figure.

**Question 7: A (7, -3), B (5, 3) and C (3, -1) are the vertices of a âˆ†ABC and AD is its median. Prove that the median AD divides âˆ†ABC into two triangles of equal areas.**

**Solution:**

So,

Area of triangle ABD:

= 1/2(7(3â€“1) + 5(1+3) + 4(â€“3â€“3))

= 1/2(14 + 20â€“24)

= 1/2(10)

= 5 sq. units â€¦(1)

Area of triangle ACD:

= 1/2(7(â€“1â€“1) + 3(1 + 3) + 4(â€“3 + 1))

= 1/2(â€“14 + 12â€“8)

= 1/2(10)

= 5 sq. units â€¦.(2)

From (1) and (2), we conclude that Area of triangle ABD and ACD is equal.

Hence proved.

**Question 8: Find the area of âˆ†ABC with A (1, -4) and midpoints of sides through A being (2, -1) and (0, -1). **

**Solution:**

Given: A âˆ†ABC with A(1, -4)

**Question 9: A(6, 1), B(8, 2) and C(9, 4) are the vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of âˆ†ADE. **

**Solution:**

A(6, 1), B(8, 2) and C(9, 4) are the three vertices of a parallelogram ABCD.

E is the midpoint of DC.

Join AE, AC and BD which intersects at O, where O is midpoint of AC.

**Question 10: (i) If the vertices of âˆ†ABC be A (1, -3), B (4, p) and C (-9, 7) and its area is 15 square units, find the values of p. **

**(ii) The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is **

**(7/2, y), find the value of y. **

**Solution**:

We know that:

**Question 11: Find the value of k so that the area of the triangle with vertices A (k + 1, 1), B (4, -3) and C (7, -k) is 6 square units. **

**Solution**:

**Question 12: For what value of k (k > 0) is the area of the triangle with vertices (-2, 5), (k, -4) and (2k + 1, 10) equal to 53 square units? **

**Solution**:

**Question 13: Show that the following points are collinear.**

**(i) A (2, -2), B (-3, 8) and C (-1, 4)**

**(ii) A (-5, 1), B (5, 5) and C (10, 7)**

**(iii) A (5, 1), B (1, -1) and C (11, 4)**

**(iv) A (8, 1), B (3, -4) and C (2, -5)**

**Solution:**

Points are collinear if the area of a triangle is equal to zero.

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)

(i) A (2, -2), B (-3, 8) and C (-1, 4)

Î” = 1/2{2 (8â€“ 4) + (â€“3) (4 + 2) â€“1 (2â€“ 8)}

Î” = 1/2 {8â€“18 + 10}

Î” = 0

Hence points are collinear.

(ii) A (-5, 1), B (5, 5) and C (10, 7)

Î” = 1/2{â€“5(5â€“ 7) + 5 (7â€“1) + 10 (1â€“5)}

Î” = 1/2{10 + 30â€“40}

Î” = 0

Hence points are collinear.

(iii) A (5, 1), B (1, -1) and C (11, 4)

Î” = 1/2{5(â€“1â€“ 4) + 1 (4â€“ 1) + 11 (1 + 1)}

= 1/2{â€“25 + 3 + 22}

= 0

Hence points are collinear.

(iv) A (8, 1), B (3, -4) and C (2, -5)

Î” = 1/2{8(â€“4 + 5) + 3 (â€“5â€“1) + 2 (1 + 4)}

= 1/2{8â€“18 + 10}

= 0

Hence points are collinear.

**Question 14: Find the value of x for which the points A (x, 2), B (-3, -4) and C (7, -5) are collinear. **

**Solution:**

Points are A (x, 2), B (-3, -4) and C (7, -5) are collinear.

Which means area of triangle ABC = 0

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)Since points are collinear:

Â½(x + 63) = 0

Or x = -63

**Question 15: For what value of x are the points A (-3, 12), B (7, 6) and C (x, 9) collinear?**

**Solution:**

Points are A (-3, 12), B (7, 6) and C (x, 9) are collinear.

Which means area of triangle ABC = 0

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)Since points are collinear:

Â½(6x -12) = 0

Or x = 2

**Question 16: For what value of y are points P (1, 4), Q (3, y) and R (-3, 16) are collinear?**

**Solution:**

Points are P (1, 4), Q (3, y) and R (-3, 16) are collinear.

Which means area of triangle PQR = 0

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)Since points are collinear:

Â½(4y + 8) = 0

Or y = -2

**Question 17: Find the value of y for which the points A (-3, 9), B (2, y) and C (4, -5) are collinear.**

**Solution:**

Points are A (-3, 9), B (2, y) and C (4, -5) are collinear.

Which means area of triangle ABC = 0

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)Since points are collinear:

Â½(-7y – 7) = 0

Or y = -1

**Question 18: For what values of k are the points A (8, 1), B (3, -2k) and C (k, -5) collinear. **

**Solution:**

Points are A (8, 1), B (3, -2k) and C (k, -5) are collinear.

Which means area of triangle ABC = 0

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)Since points are collinear:

Â½(2k^2 – 15k + 22) = 0

Or 2k^2 – 15k + 22 = 0

2k^2 – 11k – 4k + 22 = 0

k(2k – 11) – 2(2k – 11) = 0

(k-2)(2k -11) = 0

k = 2 or k = 11/2. Answer.

**Question 19: Find a relation between x and y, if the points A(2, 1), B(x, y) and C(7, 5) are collinear. **

**Solution: **

Points are A(2, 1), B(x, y) and C(7, 5) are collinear.

Which means area of triangle ABC = 0

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)Since points are collinear:

Â½(4x -5y – 3) = 0

4x – 5y – 3 = 0

Relationship between x and y.

**Question 20: Find a relation between x and y, if the points A(x, y), B(-5, 7) and C(-4, 5) are collinear. **

**Solution: **

Points are A(x, y), B(-5, 7) and C(-4, 5) are collinear.

Which means area of triangle ABC = 0

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)Since points are collinear:

Â½(2x +y + 3) = 0

2x +y + 3 = 0

Relationship between x and y.

**Question 21: Prove that the points A(a, 0), B(0, b) and C(1, 1) are collinear, if 1/a + 1/b = 1.**

**Solution: **

Points are A(a, 0), B(0, b) and C(1, 1) are collinear.

Which means area of triangle ABC = 0

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)= Â½[(a(b-1) + 0(1-0) + 1(0-b)]

= Â½[ab â€“ a + 0 â€“ b]

Since points are collinear:

Â½(ab -a-b) = 0

ab – a – b = 0

Divide each term by “ab”, we get

1 – 1/b -1/a = 0

or 1/a + 1/b = 1. hence proved.

**Question 22: If the points P(-3, 9), Q(a, b) and R(4, -5) are collinear and a + b = 1, find the values of a and b. **

**Solution: **

Points are P(-3, 9), Q(a, b) and R(4, -5) are collinear.

Which means area of triangle = 0

\(Area of \bigtriangleup ABC=\frac{1}{2}\left [ x_{1}\left ( y_{2} – y_{3}\right ) + x_{2} \left( y_{3} – y_{1}\right )+ x_{3}\left( y_{1} – y_{2}\right )\right ]\)Since points are collinear, we have

Â½(-14a – 7b + 21) = 0

-14a – 7b + 21 = 0

2a + b =3 …..(1)

a + b = 1 …..(2) (given)

From (1) and (2)

a = 2 and b = -1

**Question 23: Find the area of âˆ†ABC with vertices A (0, -1), B (2, 1) and C (0, 3). Also, find the area of the triangle formed by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is 4 : 1.**

**Solution:**

Vertices of âˆ†ABC are A (0, -1), B (2, 1) and C (0, 3)

**Area of triangle DEF:**

= 1/2[1+0+1]

= 1 sq. units

Therefore,

Ratio in the area of triangles ABC and DEF = 4/1 = 4:1.

**Question 24: If a â‰ b â‰ c, prove that (a, aÂ²), (b, bÂ²), (0, 0) will not be collinear. **

**Solution:**

Let A (a, aÂ²), B (b, bÂ²) and C (0, 0) are the vertices of a triangle.

Let us assume that that points are collinear, then area of âˆ†ABC must be zero.

Now, area of âˆ†ABC

Which is contraction to our assumption.

This implies points are not be collinear. Hence proved.

**R S Aggarwal Solutions for Class 10 Maths Chapter 6 Coordinate Geometry Exercise 6C Topics**

Class 10 Maths Chapter 6 Coordinate Geometry Exercise 6C is based on following topics:

**Area of a triangle**

Area of âˆ†ABC whose vertices are is (x_1,y_1 ),(x_2,y_2 )and (x_3,y_3 ) are

Students can also find the collinearity of three points using area of triangle formula.